Question

1. A 12 m beam is simply supported roller at one end, and hinged at the other end. The beam has constant (EI) and the load on the beam is illustrated as shown in Figure 1. Using the finite difference method, determine deflections in the beam. Take h=∆x = ⁄ .
   9. = 40 kN/m 50 mm 50 mm 6m Constant EIM 12 m Figure 1
   

Answer 1

ANSWER :
ke Now, RA= Rp = Total load 2 [L 0+40 2 0x6]x2 2 (i symetric toading) RA= R₂ = 120KN Now, consider a section x-x X distance 501 at o loading 6m = 400 kalm no c) kallm. C 40 x)to) BMX = RAK i-am = х 2 = RAK- 20 x 3 18 BMA=0 3 BMD = 120x6- 20 18 8 XG BMDE = 480 kn-m 3 20 3 330 kNomos BMc = 120x3x 18

@ we have given Дх 12 4 3x = 3ml So, there will be 5 nodes. 40 knlm. Loading Diagram B D 22 IE 480 kNm. 330KNM 330kNm. BMD D B 480 EI 330 ет. 330 ет . Diag EI А. c D E B Nodal Diagram in deflections are 4,42,43 Yu and Чs

(cu At A and B i deflection will be zero=0 -0 :. 4, -45 and at c and e both are symmetric so, Y2 = 44 Now, Apply finite Difference method dy MA Yox - 2y + Intl Ein Cox2 dx2 At Node 2, M₂ EI y-2y + Y3 Tox) 330 0-242 +2 GI 32 0 2970 EI At Node Mz EI ₂₂ - 243 +45 (PR) 2 242-1843 480 СТ 32 4320 242 24₃ EI Now compare egn egn 6 +

get, -Y 2 = 7290 GI (-ve sigri indicates deflection is opposite to BMD) 1-42 = -44 = 5130 EI Iinal Answers: 4,=0 Y. 5130 (W) EI 7290 or EI ( 90' = 5130 EI %=0